# How do you solve 3|2x + -5| + 3 = 42?

Mar 18, 2018

$x = 9 \mathmr{and} x = - 4$

#### Explanation:

Here, $3 | 2 x + \left(- 5\right) | + 3 = 42 ,$

Adding $\left(- 3\right)$ both sides

$3 | 2 x + \left(- 5\right) | + 3 + \left(- 3\right) = 42 + \left(- 3\right)$

$\implies 3 | 2 x + \left(- 5\right) | = 39 ,$

Dividing both sides by 3

$\frac{\cancel{3} | 2 x + \left(- 5\right) |}{\cancel{3}} = \frac{39}{3} \implies | 2 x + \left(- 5\right) | = 13$

$2 x + \left(- 5\right) = 13 \mathmr{and} 2 x + \left(- 5\right) = - 13$

Adding $5$ both sides

$2 x + \left(- 5\right) + 5 = 13 + 5 \mathmr{and} 2 x + \left(- 5\right) + 5 = - 13 + 5$

$\implies 2 x = 18 \mathmr{and} 2 x = - 8 \implies x = 9 \mathmr{and} x = - 4$

Mar 18, 2018

$x = - 4 , 9$

#### Explanation:

Solve:

$3 \left\mid 2 x + - 5 \right\mid + 3 = 42$

Simplify the absolute value.

$3 \left\mid 2 x - 5 \right\mid + 3 = 42$

Subtract $3$ from both sides of the equation.

$3 \left\mid 2 x - 5 \right\mid + 3 - 3 = 42 - 3$

Simplify.

$3 \left\mid 2 x - 5 \right\mid + 0 = 39$

$3 \left\mid 2 x - 5 \right\mid = 39$

Divide both sides by $3$.

$\frac{{\cancel{3}}^{1} \left\mid 2 x - 5 \right\mid}{\cancel{3}} ^ 1 = {\cancel{39}}^{13} / {\cancel{3}}^{1}$

Simplify.

$\left\mid 2 x - 5 \right\mid = 13$

Since $\left\mid \pm a \right\mid = a$, we can break the equation into two equations:

$2 x - 5 = 13$ and $- \left(2 x - 5\right) = 13$

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Solve the first equation:

Add $5$ to both sides of the equation.

$2 x - 5 + 5 = 13 + 5$

Simplify.

$2 x = 18$

Divide both sides by $2$.

$\frac{{\cancel{2}}^{1} x}{\cancel{2}} ^ 1 = {\cancel{18}}^{9} / {\cancel{2}}^{1}$

Simplify.

$x = 9$

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Solve the second equation:

$- \left(2 x - 5\right) = 13$

Expand.

$- 2 x + 5 = 13$

Subtract $5$ from both sides.

$- 2 x + 5 - 5 = 13 - 5$

Simplify.

$- 2 x + 0 = 8$

$- 2 x = 8$

Divide both sides by $- 2$.

(cancel(-2)^1x)/(cancel(-2)^1)=cancel8^4/(cancel(-2)^1

Simplify.

$x = - 4$

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$x = - 4 , 9$